Definition: We say that $f(x)\neq \text{ scalar } \in F[x]$ is irreducible if for any $g(x)\in F[x]$ such that $g(x)\mid f(x),$ either $g(x)=\alpha$ is scalar or $g(x)=\alpha \cdot f(x)$ for a scalar $\alpha \neq 0$

Definiton: $g(x) \text{ divides } f(x),\ g(x) \mid f(x) \text{ if } \exists \ q(x)\in F[x]\ st. \ f(x)=g(x)\cdot q(x)$

We say $g(x)$ is a proper divisor of $f(x)$ if $g(x)\mid f(x)$ and $g(x)\neq \text{scalar and }g(x)\neq \alpha \cdot f(x),\ \alpha \in \R$

Example:

$f(x)=x^2 -1 \in \R[x]$

• $\underbrace{(\frac{1}{2}x^2+\frac{1}{2})}_{g}\underbrace{(-2)}_h=\underbrace{x^2-1}_f\$ $g,h$ are not proper ideals
• $x^2 -1=\underbrace{(x-1)}_p\underbrace{(x-1)}_q\$ $p,\ q$ are proper divisors

Remark: $g\mid f$ is a proper divisor $\iff 0<\text{deg(g)}< \text{deg(f)}$

Proof:

$\impliedby 0 <\text{deg(g)} <\text{deg(f)}\implies g\neq \alpha \cdot f\text{ as } (\alpha \cdot f)=\text{deg(f)}$

$\implies \text{deg(g)} > 0$ ← $g$ is a proper divisor

If $\text{deg(g)=deg(f)},$ then$f(x)=g(x)\cdot q(x)\implies \text{deg(q(x))}=0 \implies q(x)=\alpha \in \R$

This is a contradiction since $g$ is a proper divisor $\square$

Dictionary:

$I=\braket{\ f(x) \ }\lhd F[x],\ J=\braket{\ g(x)\ }\lhd F[x]$

$$ \text{Dictionary} \\ \def\arraystretch{1.7} \begin{array}{c | c} I\rhd R & F[x] \\ \hline I=0 & f(x)=0 \\ \hline I=R & f(x) \text{ is a non-zero scalar} \\ \hline I\lhd R\text{ (proper) } & \text{ deg( f(x) ) > 0} \\ \hline I\subseteq J & g(x) \mid f(x) \\ \hline I= J & g(x) =\alpha \cdot f(x),\alpha \in \R,\alpha \neq 0 \\ \hline I\sub J & g(x) \mid f(x)\text{, but }g(x)\neq \alpha \cdot f(x) \\ \hline I\sub J \lhd R \text{ (proper) } & g(x) \mid f(x),g(x)\neq \alpha \cdot f(x), g(x)\text{ not a scalar }\newline &\iff g(x) \text{ is a proper divisor of } f(x) \\ \hline I\lhd R \text{ (maximal) } & f(x)\text{ has no proper divisors (irreducible)} \\ \end{array} $$

Corollary:

TFAE: $(0\neq f(x)\in F[x]$

1. $f(x)$ is irreducible